Bài 2:
Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
\(\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)
\(\Leftrightarrow\frac{2x^2}{4}=\frac{3y^2}{27}=\frac{z^2}{25}\)
Ta lại có: \(2x^2+3y^2-z^2=20\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{2x^2}{4}=\frac{3y^2}{27}=\frac{z^2}{25}=\frac{2x^2+3y^2-z^2}{4+27-25}=\frac{20}{6}=\frac{10}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2x^2}{4}=\frac{10}{3}\\\frac{3y^2}{27}=\frac{10}{3}\\\frac{z^2}{25}=\frac{10}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2=\frac{10\cdot4}{3}=\frac{40}{3}\\3y^2=\frac{270}{3}=90\\z^2=\frac{250}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\frac{20}{3}\\y^2=30\\z^2=\frac{250}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\frac{2\sqrt{5}}{\sqrt{3}}\\y=\pm\sqrt{30}\\z=\pm\frac{5\sqrt{10}}{\sqrt{3}}\end{matrix}\right.\)
